I am trying to generate models for the following sequent.

$$\exists x \exists y \forall z (z = x \lor z = y)$$

What I have come up with is this.

$$A= \{0,1\}$$

So in this model, for all the values of $z$ there exists $x$ such that $z = x$ or there exists $y$ such that $z = y$. Is this model correct? What other models are there that satisfy this?


You should be careful with order of quantifiers. You are given a predicate $$\exists x \exists y \forall z (z = x ∨ z = y)$$ which is read as $$\text{ there is } x \text { such that there is } y \text { for all } z \dots $$ while you read it as $$ \text { for all the values of } z \text{ there exists } x \dots$$ which would correspond to $$\forall z \exists x \dots$$ When you change the order of quantifiers the meaning (and hence the truth value) of the sentence changes. For example consider the set of natural numbers excluding 0. Then $$\forall x \exists y (x \text { divides } y)$$ is a true sentence since for any $x$ take $y=x$. But the sentence $$\exists x \forall y (x \text { divides } y)$$ is false since $x$ does not divide $x+1$.

Now as for your question, yes $A=\{0, 1\}$ models this sentence. In fact you could also take a set consisting of a single element, for instance $\{0\}$ or $\{1\}$ (there is no condition such as $x\neq y$).

  • $\begingroup$ Will keep that in mind .. Thank you for the detailed explanation .. It cleared my concept $\endgroup$ – Zeist Oct 7 '17 at 17:05

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